What is the gravitational potential energy of a 1.3 kg book that is 473

Gre4nikov [31]

Answer:

Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path at a constant speed, it is still accelerating, because the direction of its velocity is changing. Comment on robshowsides's post “Speed is the magnitude of velocity.

Explanation:

hope it helped tee hee

6 0

3 years ago

Read 2 more answers

Examples of uniform velocity

slava [35]

Explanation:

When a truck travels in equal distances in equal intervals of time then we say that the body has got a uniform velocity. In the above example a truck is traveling at 5 miles in all the positions at A, B, and C and all in the intervals of 5 minutes each.

8 0

3 years ago

What is the difference between a good conductor and a good insulator?

brilliants [131]

Answer:

Explanation:

In a conductor, electric current can flow freely, in an insulator it cannot.

Metals such as copper typify conductors, while most non-metallic solids are said to be good insulators, having extremely high resistance to the flow of charge through them.

Most atoms hold on to their electrons tightly and are insulators.

4 0

3 years ago

A person with mass mp = 76 kg stands on a spinning platform disk with a radius of R = 1.98 m and mass md = 191 kg. The disk is i

nalin [4]

<span>1.7 rad/s The key thing here is conservation of angular momentum. The system as a whole will retain the same angular momentum. The initial velocity is 1.7 rad/s. As the person walks closer to the center of the spinning disk, the speed will increase. But I'm not going to bother calculating by how much. Just remember the speed will increase. And then as the person walks back out to the rim to the same distance that the person originally started, the speed will decrease. But during the entire walk, the total angular momentum remained constant. And since the initial mass distribution matches the final mass distribution, the final angular speed will match the initial angular speed.</span>

3 0

4 years ago

A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh

irina1246 [14]

(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

$K=\frac{1}{2}mv^2$

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

$U=mgh$

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

$a=\frac{v-u}{t}$

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

$u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s$

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

$\Delta v = v -u$

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

$\Delta v = 0 - (+9.8 m/s)=-9.8 m/s$

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

$a=\frac{v-u}{t}$

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

$v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s$

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

$\Delta v = v -u$

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

$\Delta v = -9.8 m/s - 0=-9.8 m/s$

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

$\Delta v = v -u$

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

$\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s$

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

$F=mg$

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

$mg = ma$

which means that the acceleration is

$a= g = -9.8 m/s^2$

and the negative sign means it points downward.

7 0

3 years ago